Question

Cho M=3¹+3²+3³+...+3²⁸+3²⁹+3³⁰.

Chứng minh M chia hết cho 13

Answer 1

Ta có \(M=\left(3^1+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{28}+3^{29}+3^{30}\right)\)

\(=3\left(1+3+3^2\right)+3^4.\left(1+3+3^2\right)+...+3^{28}.\left(1+3+3^2\right)\)

\(=13\left(3+3^4+...+3^{28}\right)⋮13\Rightarrow M⋮13\)

Answer 2

M = 3¹ + 3² + 3³ +...+ 3²⁸ + 3²⁹ + 3³⁰

M = ( 3¹ + 3² + 3³) + ...+ ( 3²⁸ + 3²⁹ + 3³⁰ )

M = 3(1 + 3 + 3² ) +...+ 3²⁸( 1 + 3 + 3²)

M = 3 .13 +...+ 3²⁸.13

\(\Rightarrow M⋮13\)(đpcm)

   !!!

Answer 3

Ta có \(M=3^1+3^2+3^3+...+3^{28}+3^{29}+3^{30}\)

\(=\left(3^1+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{28}+3^{29}+3^{30}\right)\)

\(=3.\left(1+3+3^2\right)+3^4.\left(1+3+3^2\right)+...+3^{28}.\left(1+3+3^2\right)\)

\(=3.13+3^4.13+...+3^{28}.13\)

\(=13.\left(3+3^4+..+3^{28}\right)\) chia hết cho 13.

Vậy M chia hết cho 13

Answer 4

\(M=3^1+3^2+3^3+...+3^{28}+3^{29}+3^{30}\)

\(M=3^1.\left(1+3+3^2\right)+3^3.\left(1+3+3^2\right)+...+3^{28}\left(1+3+3^2\right)\)

\(M=3^1.13+3^3.13+...+3^{28}.13\)

\(M=13.\left(3^1+3^3+...+3^{28}\right)\)

\(\Rightarrow M⋮13\)